Track / Stage 4 / 122 of 340 #382 of 2000

Problem 382

AMC 12 late, AIME early

Algebra Difficulty 4.7 Find the answer

1. If $a>b>0$ , and $a+b=6 a^{\frac{1}{2}} b^{\frac{1}{2}}$ , then the value of $\frac{a^{\frac{1}{2}}+b^{\frac{1}{2}}}{a^{\frac{1}{2}}-b^{\frac{1}{2}}}$ is ( ).
(A) $\sqrt{2}$
(B) 2
(C) 3
(D) 4

Your answer

Official solution

$\begin{array}{l}\text { I. 1. A. } \\ \text { Solution: From }\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)^{2}=a+b+2 a^{\frac{1}{2}} b^{\frac{1}{2}}=8 a^{\frac{1}{2}} b^{\frac{1}{2}}, \\ \left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)^{2}=a+b-2 a^{\frac{1}{2}} b^{\frac{1}{2}}=1 u^{\frac{1}{2}} b^{\frac{1}{2}}, \\ \text { we have } \frac{a^{\frac{1}{2}}+b^{\frac{1}{2}}}{a^{\frac{1}{2}}-b^{\frac{1}{2}}}=\sqrt{\frac{13 a^{\frac{1}{2}} i^{\frac{1}{2}}}{a^{\frac{1}{2}}} b^{\frac{1}{2}}}=\sqrt{2} .\end{array}$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.